AMC 10 · 2015 · #11

Grade 8 geometry-2d
pythagorean-theoremratio-proportionarea-rectangles convert-to-algebra ↑ Prerequisites: pythagorean-theoremratio-proportion
📏 Short solution 💡 2 insights
Problem
A rectangle has its length and width in the ratio 4:3. Its diagonal has length d. The area can be written as k d² for a single fixed number k. Find that number k.

Pick an answer.

(A)
$\frac{2}{7}$
(B)
$\frac{3}{7}$
(C)
$\frac{12}{25}$
(D)
$\frac{16}{25}$
(E)
$\frac{3}{4}$

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

The size is never given, only the shape, so Tool #4 (Introduce a Variable) lets us write the sides as 4x and 3x — one unknown x carries the unknown size while locking in the 4:3 shape. Tool #1 (Draw a Diagram) shows the diagonal cutting the rectangle into a right triangle, which is the bridge from the sides to d. Then Tool #13 (Convert to Algebra) turns "area = k d²" into an equation in x; the x cancels, leaving the pure number k the problem asks for.

1STEP 1

Name the sides with one variable

Let the sides be 4x and 3x, so the 4:3 ratio holds automatically.

length=4x, width=3x
2STEP 2

Use the diagonal's right triangle

By the Pythagorean theorem, d² = (4x)² + (3x)² = 25x².

d² = (4x)² + (3x)² = 25x²
3STEP 3

Write the area and cancel x

Area = 12x² = 12/25 d², so k = 12/25 — choice (C).

area=12x² = 12·d²/25 = 12/25d² → k=12/25 (C)
Answer
12/25
Check with a concrete 4-by-3 rectangle (x=1): its area is 12 and its diagonal is √(16+9)=5, so d²=25. Then area /d² = 12/25, exactly matching k. The value is also sensible: k is less than 1/2 because a rectangle fills less than half of the square built on its diagonal, which rules out the larger options (D) 16/25 and (E) 3/4. The 25 in the denominator is the fingerprint of the 3-4-5 triangle, pointing to (C).
💡Key takeaway

Call the sides 4x and 3x; the diagonal gives d²=25x² and the area is 12x², so the area is always 12/25 of d².

  • Name the sides with one variable
  • Use the diagonal's right triangle
  • Write the area and cancel x